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Determination of the formula of Hydrated Iron(II) Sulphate Crystals

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Introduction

Determination of the formula of Hydrated Iron(II) Sulphate Crystals Hydrated FeSO4 can come in many hydrated forms. Using two methods for better accuracy and to make a comparison, I will deduce the number of water molecules per mole of a hydrated iron(II) sulphate sample. Method 1: FeSO4.nH2O(s) � FeSO4(s) + nH2O(l) =dehydrating the FeSO4 by evaporating the water and calculating the mass lost. Hazards in this experiment include the flame from the Bunsen and the possibility of the crucible shattering. By using a gentle heat for two minute intervals and the lid slightly ajar (but not so the contents might spit) the crucible shouldn't spit and the water can escape. The reactant (FeSO4) is not dangerous, unless consumed in large doses. Results: Crucible, lid and FeSO4= 21.76g FeSO4= 1.41g Heating Mass after heating(g) Overall loss(g) 1 21.10 0.66 2 21.10 0.66 3 21.10 0.66 4 21.10 0.66 All four results show one and no further mass change meaning the water has been evaporated leaving just the iron(II) sulphate. Analysis: Out of the 1.41g sample 0.66g was water and the other 0.75g being FeSO4. By working out the molar ratio the relative molar mass are taken into account and the amount of water present in the hydrated FeSO4 can be determined. ...read more.

Middle

ions, in the hydrated iron(II) sulphate, into iron(III) ions: 1. Fe2+(aq) � Fe3+(aq) + e-, and 2. MnO4-(aq) + 8H+(aq) +5e- � Mn2+(aq) + 4H2O(l). To put these half equations into a full ionic equation the electrons must cancel out on either side. To do this the oxidisation equation (1) is multiplied by 5 and combined with the reduction equation (2) to give the complete redox ionic equation: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) � Mn2+(aq) + 5Fe3+(aq) + 4H2O(l). From this equation the molar ratio between potassium permanganate and iron(II) sulphate is shown to be 1:5, so (2.2275�10-4)�(5) moles of hydrated iron(II) sulphate were present in 25cm3, and (2.2275�10-3) � (5) were present overall in the whole 250cm3 solution. Moles of iron(II) sulphate = (2.2275�10-3) � (5) = 1.11375�10-2 moles. The mass of the iron(II) sulphate that reacted was moles � molecular mass; (1.11375�10-2) � (151.9) = 1.69g.(3.s.f) As the water in the hydrated iron(II) sulphate weighed at the start played no part in the reaction, subtracting the mass of the anhydrous iron(II) sulphate from the initial hydrated form will give the mass of water present in the sample; 3.05g - 1.69g = 1.36g of water. ...read more.

Conclusion

This method generally relies on chemistry instead of maths and the only variable that could have caused the off answer was again the scales. Carrying out this experiment on a larger scale would be far too dangerous in a school due to the use of fire. The problem with method one being more in touch with chemistry is the risk of procedural errors. On heating the melanterite the substance changed in colour from blue to white indicating the water had been eliminated. However there were also areas of dark red which indicates a further reaction. The heat was too strong and iron(III) oxide was formed with sulphur di- and trioxide: 2FeSO4(s) � Fe2O3(s) + SO2(g) + SO3(g). These gases also escaped with the water and gave the impression that there was more water present as some sulphur and oxygen escaped as well. Heating the melanterite more gently and patiently for less time would make sure this didn't happen. If this didn't happen I am confident that this method would be more accurate than a titration as it is more purely chemistry instead of maths, which holds a greater chance of carrying errors over, despite my results suggesting otherwise. ?? ?? ?? ?? Rob H ...read more.

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